- #1

- 24

- 0

## Homework Statement

Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge -2.00μC of mass 6.00g is initially located 3.00 cm from each of these charges and released from rest. You observe that the initial acceleration of -2.00μC is 354m/s^2 upward, parallel to the line connecting the two point charges. Find q1 and q2.

## Homework Equations

## F=ma ##

## F= kQq1/r^2 ##

## The Attempt at a Solution

I drew a sketch of the problem, with q1 directly above q2 (.045m away). The first thing I did was find the total force by plugging in the mass and acceleration:

## F = (0.006kg)*(354m/s^2) = 2.124 N ##

Next, I saw that since the object is accelerating upward and parallel to the line connecting the two point charges, then q1 must be positive and q2 must be negative. Since Q is moving straight up, there is no movement in the x direction.

With that, I figured that:

## F = Fy ##

## 2.124 = 2 FQy ##

## FQy = 1.062 ##

So far, I get that the total force is the vector sum of Fq1y and Fq2y. And since these vectors are in the y-direction, you can add their magnitudes (I think). I tried solving for q1 in Coulombs' law:

## F*r^2/kQ = q1 ##

## (1.062)(.03^2)/(8.99*10^9)(2*10^-6) = 5.3*10^-8 C ##

(That should be 10^-6 but latex keeps putting the - sign up and not the 6. Not sure how to fix that but I calculated it correctly at that part)

But this isn't the right answer. I'm pretty sure that q1 and q2 have the same magnitude, just different charges (q1 is + and q2 is -), but I don't know what to do in order to find them. Any help would be appreciated. Thanks!